Cos/1+sin + 1+sin/cos = 2sec

Question

I thought it could possible be solved by a double angle formula, but I cant seem to make it work

Answer 1

   cos          (1+sin) 

(1+sin)  +     cos

 

=   cos² + (1+sin)² 

         cos*(1+sin)

 

=   cos² + 1 + 2sin + sin² 

           cos*(1+sin)

 

we know that cos² + sin² = 1

 

=    (1+1+2sin) 

      cos*(1+sin) 

 

=   (2+2sin)

    cos*(1+sin)

 

=      2(1+sin)

      cos*(1+sin)

 

=    2   

    cos 

 

=  2sec

Answer 2

I assume the grouping is

 

cos/(1 + sin) + (1 + sin)/cos = 2 sec      common denominator is cos(1 + sin)

[ cos · cos + (1 + sin)(1 + sin) ] / [cos(1 + sin)] = 2 sec 

[ cos² + sin² + 2 sin + 1 ] / [cos(1 + sin)] = 2 sec

[ 1 + 2 sin + 1] /  [cos(1 + sin)] = 2 sec

[ 2 sin + 2] / [cos(1 + sin)] = 2 sec

[ 2(sin + 1] / [cos(1 + sin)] = 2 sec

2 1/cos = 2 sec

2 sec = 2sec

Answer 3

Hannah,

 

You can't make it work because it's impossible given that your equality above is false.

 

If true, it should hold for any angle, right?  So take 30^o and evaluate the left and right hand sides and see if they match.

 

Cos/1+sin + 1+sin/cos = 2sec , and cos = 0.866025, sin = 0.500, tan = sin/cos = 0.57735, and sec = 1/cos = 1.15470

 

Then,

0.866025 + 0.50 + 1.0 + 0.57735  =?  2(1.15470)

2.943375  =?  2.30940

 

Obviously, no match, so relationship is false.  Did you make a mistake in typing it?